$\begin{cases}b(1)=16\\\\ b(n)=b(n-1)+1 \end{cases}$ Find the $2^{\text{nd}}$ term in the sequence.
This is a recursive formula. It tells us that the first term is $16$ and that the common difference is $1$. $\begin{aligned} {b(1)}&=16 \\\\ {b(2)}&={b(1)}+1=17 \end{aligned}$ The $2^{\text{nd}}$ term is $17$.